Question

find the
Half range sine series and cosine series of given function f(x) =x, 0<x<2​

Answer 1

Answer:In this lecture we consider the Fourier Expansions for Even and Odd functions, which give rise to cosine and sine half

range Fourier Expansions. If we are only given values of a function f(x) over half of the range [0, L], we can define two

different extensions of f to the full range [−L, L], which yield distinct Fourier Expansions. The even extension gives rise

to a half range cosine series, while the odd extension gives rise to a half range sine series.

Key Concepts: Even and Odd Functions; Half Range Fourier Expansions; Even and Odd Extensions

Step-by-step explanation:

Answer 2

Answer:

The Half range sine series and cosine series is $f(x) = - \frac{4}{\pi }$∑(n =1 to ∞) $\frac{(-1)^{n} }{n} sin(\frac{n\pi x}{2} )$

Step-by-step explanation:

Given:  The function f(x) =x, 0<x<2​

To find: The Half range sine series and cosine series

Solution:

Let,

f(x) =x, 0< x < 2

Here, L = 2

Half range sine and cosine series:

A function can be stretched into a sequence of sine terms or cosine terms if it is specified over half the range, such as 0 to L, rather than the whole range from L to L.

Suppose the Half range sine series of f(x) is,

$f(x) =$∑(n =1 to ∞)$b_{n} sin(\frac{n\pi x}{L} )$

f(x) = ∑(n =1 to ∞) $b_{n} sin(\frac{n\pi x}{2} )$     Where L = 2

$b_{n} = \frac{2}{2} \int\limits^2_0 {f(x)sin(\frac{n\pi x}{2} } \, dx$

  = $\int\limits^0_2 {xsin(\frac{\pi nx}{2} )} \, dx$

= [$x(\frac{2}{n\pi } cos(\frac{n\pi x}{2} )) - 1 (-\frac{4}{n^{2} \pi ^{2} } )sin(\frac{n\pi x}{2} )$](0 to 2)

= { - $\frac{4}{n\pi } (-1)^{n} - 0$} - {0 - 0 }

$b_{n} = - \frac{4}{n\pi } (-1)^{n}$

$f(x) = - \frac{4}{\pi }$∑(n =1 to ∞) $\frac{(-1)^{n} }{n} sin(\frac{n\pi x}{2} )$

Final answer:

The Half range sine series and cosine series is $f(x) = - \frac{4}{\pi }$∑(n =1 to ∞) $\frac{(-1)^{n} }{n} sin(\frac{n\pi x}{2} )$

SPJ3